How many valid host addresses will be unused on the three point-to-point links combined if VLSM is not used?

[S9539245H]

In the network shown in the graphic, three bits were borrowed from the host portion of a Class C address. How many valid host addresses will be unused on the three point-to-point links combined if VLSM is not used?
	3
	4
	12
	36
	84
	180

I need someone to explain this to me, thanks!

[Gabriel Hill]

Hello Jason,

So basically you have a class C network that has had 3 bits borrowed from the host portion.

255.255.255.0 = regular class C = /24

255.255.255.224 = 3 bits stolen from host portion =/27

Therefore if the three point to point links are not using VLSM, they will each use a /27.

So for example:

192.168.1.0 /27

192.168.1.32 /27

192.168.1.64 /27

Each network can have 30 hosts on it, 2 ^ 5 (5 bits left in the subnet after 3 were taken away) = 32 - (subnet IP and broadcast IP) = 30.

We have a total of 90 usable IP's between the 3 networks. We have 6 point-to-point links, each will need its own IP.

90 - 6 = 84

Hope that makes sense.

-Gabriel

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